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Find the value of $k$ such that $x + 2$ is a factor of $2x^3 + x^2 + k x + 6$ then factorise completely.

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    $\begingroup$ Dear @JoeyZou, your edit is probably among the most useless edits ever. First, it is performed on a 2 years old post; second, it only makes a minor change, and not related to mathematics; third, the use of the short infinitive to construct the subjunctive was correct, and your use of the indicative mood is, in fact, mistaken. Please think about this. $\endgroup$
    – Alex M.
    Sep 1, 2016 at 20:33
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    $\begingroup$ @AlexM. Please take a look at the edit history. My edit was an "improvement" on another user's suggested edit made about 30 minutes ago, which removed an image present in the question and replaced it with text. Thank you for bringing the short infinitive to my attention--I was not aware that was grammatically correct. Apologies that my attempt at "correcting" what I thought was an otherwise valid edit is what led you to think that I capriciously edited one word on a two-year post. $\endgroup$
    – Joey Zou
    Sep 1, 2016 at 20:49
  • $\begingroup$ Dear @BD., please stop making useless edits (and, worse, on inactive old posts). $\endgroup$
    – Alex M.
    Sep 1, 2016 at 21:29

4 Answers 4

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Hint: Let $2x^3+x^2+kx+6 = (x+3)P(x)$ for some quadratic $P(x)$. Now, plug in $x = -3$.

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  • $\begingroup$ I understand that a value of x would be -3 if the expression 2x^3 + x^2 + kx + 6 equated to 0. The question gives an expression and not an equation, so can't 2x^3 + x^2 + kx + 6 be equal to any number? In that case how would we figure k out? $\endgroup$ Aug 17, 2014 at 7:36
  • $\begingroup$ It is the other way around, if x = -3 then expression = 0, not if expression = 0, x has to be -3. Then you can just fill in -3 for x. Just like in my other answer. (That you should accept as answer ;) ) Just for thought: Expressions and equations aren't all that different. $\endgroup$
    – Pieter21
    Aug 17, 2014 at 7:44
  • $\begingroup$ We just wrote an equation, specifically, $2x^3+x^2+kx+6 = (x+3)P(x)$. Without doing more work, $P(x)$ could be anything, but that's ok. When you plug in $x = -3$, you get $0$ on the right side, and something involving $k$ on the left side. That's an equation which you can use to solve for $k$. $\endgroup$
    – JimmyK4542
    Aug 17, 2014 at 13:18
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If $(x+3)$ is a factor, this function should evaluate to $0$ for $x = -3$.

Fill in $-3$ for $x$ and have $-54 + 9 - 3k + 6 = 0$, and find $k = 13$

Did I say $k = 13$, that was a bad copying to or from my notes?

$-54 + 9 + 6 = -39 = 3k $

so $k$ should have been $-13$.

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  • $\begingroup$ Why should it evaluate to 0? I thought that 2x^3 + x^2 + kx + 6 could be any number and did not necessarily have to equal to 0. The question gives an expression, not an equation, so can't 2x^3 + x^2 + kx + 6 be any number other than 0? $\endgroup$ Aug 17, 2014 at 7:34
  • $\begingroup$ It can be any number, except when x = -3. Because (x + 3) is a factor and (-3 + 3) * something will result in 0. $\endgroup$
    – Pieter21
    Aug 17, 2014 at 7:38
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By long division we can obtain polynomials $q(x)$ and $r(x)$ such that : $2x^3+x^2+kx+6 = (x+3)q(x) + r(x)$

$r(x) = 0$ since $(x+3)$ is a factor :

$2x^3+x^2+kx+6 = (x+3)q(x) + 0$

This is an equality and must work for ALL values of x, including x = -3 :

$2(-3)^3+(-3)^2+k(-3)+6 = (-3+3)q(-3)$

$2(-3)^3+(-3)^2+k(-3)+6 = 0$

$k$ can be solved

You may refer factor theorem for more info

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Since, $x+3$ is a factor of the equation, then $f(-3)=0$

submit $x=-3$ to get $k$.

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